Proving Biconditional Equivalence

[maxsthekat]

Homework Statement

I have to prove that !(p <-> q) = (p <-> !q)


2. The attempt at a solution
I started by trying to just work out what each side of the equation was. So, starting with the left hand side
!(p <-> q) = !((p->q) * (q->p)) (biconditional law)
= !((!p + q) * (!q + p)) (implication law)
= !(!p + q) + !(!q + p) (DeMorgan's theorem)
= (p * !q) + (q * !p) (DeMorgan's theorem)

Then, for the right hand side, I tried doing similarly:
(p <-> !q) = (p -> !q) * (!q -> p) (biconditional law)
= (!p + !q) * (q + p) (implication law)

But, here I get stuck... I've thought about applying DeMorgan's, but it seems like the two expressions won't be equivalent if I do this. I've verified this equivalence is true using truth tables, but the equivalence logic seems to be escaping me. Can anyone lend a hand?

Thanks! :)

-Max

 

[JSuarez]

A truth table would be much faster, but if you don't want that, you-ll have to develop (!p + !q) * (q + p), using the distributivity of *.